Lab Values for Nurses: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Lab values & electrolytes

Method

  1. Compare the value with the laboratory’s own reference range, then with the patient’s baseline and trend.
  2. Check the sample: hemolysis, a delay, or blood drawn near an IV can give false results.
  3. Link the value to its typical signs and to the drugs and conditions that cause it.
  4. Look for interactions: low potassium or magnesium increases digoxin toxicity; dehydration raises lithium.
  5. Act on the patient: assess, hold a drug if the level is unsafe, read back critical values and inform the prescriber.

Arterial blood gas interpretation

Normal ranges
pH 7.35–7.45, PaCO₂ 35–45 mmHg (4.7–6.0 kPa), HCO₃⁻ 22–26 mmol/L
ROME
Respiratory: PaCO₂ opposite to pH; Metabolic: HCO₃⁻ equal to pH
Winter’s formula
PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2
Metabolic alkalosis
PaCO₂ ≈ 0.7·HCO₃⁻ + 21
Anion gap
AG = Na⁺ − (Cl⁻ + HCO₃⁻) ≈ 8–12
Albumin-corrected gap
AG + 2.5·(4.0 − alb g/dL) = AG + 0.25·(40 − alb g/L)
Delta ratio
(AG − 12)/(24 − HCO₃⁻)
P/F ratio
PaO₂/FiO₂
Alveolar gas equation
PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8
Serum osmolality
2·Na⁺ + glucose + urea (2·Na⁺ + glucose/18 + BUN/2.8)

Method

  1. Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
  2. Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
  3. Look at the other value: normal means uncompensated, abnormal in the compensating direction means partial compensation, and a normal pH with both abnormal means full compensation.
  4. In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
  5. Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.

Clinical pharmacokinetics

Rate constant and half-life
k = ln 2/t½ = 0.693/t½
First-order decay
C = C₀·e−kt = C₀·(1/2)t/t½
Two levels
k = ln(C₁/C₂)/Δt
Volume and clearance
Vd = dose/C₀, CL = k·Vd
Loading dose
LD = Vd·C/(S·F)
Maintenance rate
R = CL·Css/(S·F)
Average steady state
Css,avg = S·F·D/(CL·τ)
Steady state
1 − (1/2)ⁿ after n half-lives; ≈ 97% after 5
Infusion
C = (R/CL)(1 − e−kt)
Interval and dose from peak and trough
τ = ln(Cpeak/Ctrough)/k, D = Vd·(Cpeak − Ctrough)
Cockcroft–Gault
CrCl = (140 − age) × kg/(72 × SCr mg/dL), × 0.85 for women
Renal adjustment
dose × CrCl/normal; Dettli Q = 1 − fe·(1 − CrCl/normal)

Method

  1. Convert the half-life to a rate constant: k = 0.693/t½.
  2. Turn per-kg values into totals: Vd (L) = L/kg × weight, CL (L/h) = L/kg/h × weight.
  3. Pick the relation: C = C₀·e−kt, LD = Vd·C/(S·F), maintenance rate = CL·Css/(S·F), τ = ln(Cpeak/Ctrough)/k.
  4. For kidney function, estimate CrCl with Cockcroft–Gault first, then scale the dose or the interval.
  5. Check the result: levels fall by half every half-life, and steady state takes about five half-lives.

2Practice problems

1.Lab values & electrolytesMedium

When should a blood sample for a digoxin level be drawn to be interpreted correctly?

  1. AAt least 6–8 hours after a dose, often just before the next one
  2. BAt any time, because digoxin levels do not change during the day
  3. CWithin 30 minutes after an oral dose to show the peak level
  4. DOnly after a meal, because food is needed for absorption
2.Lab values & electrolytesHard

A patient with heart failure takes lisinopril, spironolactone and a potassium supplement. Today the potassium is 6.0 mEq/L. What is the most likely explanation?

  1. ASpironolactone causes potassium loss
  2. BSeveral medicines that raise potassium are being taken together
  3. CLisinopril lowers potassium
  4. DPotassium supplements have no effect on the serum level
3.Arterial blood gas interpretationMedium
pH7.357.457.62PaCO₂mmHg354524HCO₃⁻mmol/L222624uncompensatedpartially compensatedfully compensated

A practice arterial blood gas shows pH 7.62, PaCO₂ 24 mmHg (3.2 kPa) and HCO₃⁻ 24 mmol/L. The primary disorder is respiratory alkalosis. How far is it compensated? Click your answer.

pH 7.62PaCO₂ 24 mmHg (3.2 kPa)HCO₃⁻ 24 mmol/L
  1. 1uncompensated
  2. 2partially compensated
  3. 3fully compensated
4.Arterial blood gas interpretationHard

Practice values: Na⁺ 141, Cl⁻ 114, HCO₃⁻ 12 mmol/L and albumin 15 g/L. Find the albumin-corrected anion gap AGc = AG + 0.25·(40 − albumin).

Na⁺ 141 mmol/LCl⁻ 114 mmol/LHCO₃⁻ 12 mmol/Lalbumin 15 g/L
AGc = mmol/L
5.Clinical pharmacokineticsMedium
0.1110t (h)C (mg/L)1 h6 mg/L7 h1.1 mg/L

Practice exercise (made-up values, not clinical guidance). After an IV dose of tobramycin, the level is 6 mg/L at 1 h and 1.1 mg/L at 7 h. Assume first-order elimination. Find the elimination rate constant k.

C1 = 6 mg/L (1 h)C2 = 1.1 mg/L (7 h)
k = h⁻¹
6.Clinical pharmacokineticsHard
Vd = 0.7 L/kg8 → 20 mg/LLD = ? mg80 kg · S = 0.92 · F = 1

Practice exercise (made-up values, not clinical guidance). An 80 kg patient has a phenytoin level of 8 mg/L; the target is 20 mg/L. The drug is given intravenously as phenytoin sodium (S = 0.92), and the volume of distribution is 0.7 L/kg. What extra loading dose LD raises the level to the target?

80 kgVd = 0.7 L/kgC = 8 → 20 mg/LS = 0.92
LD = mg
7.Clinical nursingMedium

Which statement about storing insulin is correct?

  1. AInsulin never expires once opened
  2. BInsulin should be kept in direct sunlight
  3. CUnopened insulin is kept in the fridge and must not be frozen
  4. DAll insulin should be stored in the freezer to keep it fresh
8.Clinical nursingHard

A patient has a single hot, swollen, very painful knee and a fever. Why is urgent joint aspiration important?

  1. ATo prevent osteoporosis
  2. BTo rule out septic arthritis, which can rapidly destroy the joint
  3. CTo check for rheumatoid arthritis antibodies in the joint fluid
  4. DTo measure the patient's cholesterol

3Solutions

1.Lab values & electrolytes

Answer: A · At least 6–8 hours after a dose, often just before the next one

Why this is the answer

Digoxin distributes slowly from blood to tissues, so a sample taken too soon after a dose is falsely high. Levels are drawn at least 6–8 hours after a dose, commonly as a trough before the next dose.

Answer

A · At least 6–8 hours after a dose, often just before the next one
2.Lab values & electrolytes

Answer: B · Several medicines that raise potassium are being taken together

Why this is the answer

ACE inhibitors and spironolactone both make the kidneys retain potassium, and a supplement adds more. The nurse holds the supplement, informs the prescriber and checks kidney function and the ECG.Checked against: High potassium level · MedlinePlus (NIH) (2025); Spironolactone · MedlinePlus (NIH) (2018)

Answer

B · Several medicines that raise potassium are being taken together
3.Arterial blood gas interpretation

Answer: uncompensated

Reading a blood gas (ROME): first the pH: below 7.35 is acidaemia, above 7.45 alkalaemia. Then find the value that explains it. PaCO₂ (normal 35–45 mmHg, 4.7–6.0 kPa) is the respiratory part and moves Opposite to the pH in a respiratory disorder; HCO₃⁻ (normal 22–26 mmol/L) is the metabolic part and moves in the same direction as the pH (Respiratory Opposite, Metabolic Equal). The other value shows the compensation: normal means uncompensated; abnormal in the direction that pulls the pH back while the pH is still abnormal means partially compensated; a normal pH with both values abnormal means fully compensated. Ranges differ slightly between laboratories; these are practice values.

ROME, value by value

pH 7.62: alkalaemia (above 7.45)
PaCO₂ 24 mmHg: low (35–45)
HCO₃⁻ 24 mmol/L: normal (22–26)

Compensation

uncompensated

Answer

uncompensated
4.Arterial blood gas interpretation

Answer: AGc = 21.25 mmol/L

Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.

Anion gap

AG = 141 − (114 + 12) = 15

Correction for albumin

15 + 0.25·(40 − 15) = 21.25 mmol/L

Answer

AGc = 21.25 mmol/L

Above about 12 after correction: a high anion gap.

5.Clinical pharmacokinetics

Answer: k = 0.2827 h⁻¹

Clearance and volume: at steady state the dosing rate equals the elimination rate, so the maintenance rate = CL × Css. A loading dose is LD = Vd × C/(S × F), with S the salt factor (fraction of active drug in the salt) and F the bioavailability (1 for IV). From two levels, k = ln(C₁/C₂)/Δt; for an IV bolus Vd = dose/C₀ and CL = k × Vd. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.

Elimination rate constant

k = ln(6/1.1)/6 = 1.696/6 = 0.2827 h⁻¹

Answer

k = 0.2827 h⁻¹
6.Clinical pharmacokinetics

Answer: LD = 730.4 mg

Designing a regimen: with rapid IV doses the level falls from peak to trough during the interval, so τ = ln(Cpeak/Ctrough)/k and the dose that restores the peak is Vd × (Cpeak − Ctrough). The average steady-state level is Css,avg = S·F·dose/(CL·τ). An infusion rises as C = (R/CL)(1 − e−kt). For a drug cleared by the kidneys, the dose is scaled with the creatinine clearance. The numbers are for practice only; at the bedside always follow local protocols and have high-risk doses double-checked.

Volume of distribution

Vd = 0.7·80 = 56 L

Rise needed

20 − 8 = 12 mg/L

Loading dose

LD = 56·12/(0.92·1) = 730.4 mg

Answer

LD = 730.4 mg

In practice the dose is rounded to a convenient amount.

7.Clinical nursing

Answer: C · Unopened insulin is kept in the fridge and must not be frozen

Why this is the answer

Unopened insulin is refrigerated; in-use pens or vials are usually kept at room temperature for a limited time stated by the manufacturer. Frozen or overheated insulin must be discarded.

Answer

C · Unopened insulin is kept in the fridge and must not be frozen
8.Clinical nursing

Answer: B · To rule out septic arthritis, which can rapidly destroy the joint

Why this is the answer

Septic arthritis can destroy cartilage within days. Aspiration allows microscopy and culture, and helps distinguish infection from crystal arthritis such as gout.

Answer

B · To rule out septic arthritis, which can rapidly destroy the joint

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